If the system of linear equations
$x+y+3 z=0$
$x+3 y+k^{2} z=0$
$3 x+y+3 z=0$
has a non-zero solution $(x, y, z)$ for some $k \in R$, then $x+\left(\frac{y}{z}\right)$ is equal to :
Correct Option: , 2
$x+y+3 z=0$ $\ldots \ldots(\mathrm{i})$
$x+3 y+k^{2} z=0$ $\ldots . .(\mathrm{ii})$
$3 x+y+3 z=0$ ...........(iii)
$\left|\begin{array}{ccc}1 & 1 & 3 \\ 1 & 3 & k^{2} \\ 3 & 1 & 3\end{array}\right|=0$
$\Rightarrow 9+3+3 k^{2}-27-k^{2}-3=0$
$\Rightarrow \mathrm{k}^{2}=9$
(i) $-$ (iii) $\Rightarrow-2 \mathrm{x}=0 \Rightarrow \mathrm{x}=0$
Now from $(\mathrm{i}) \Rightarrow \mathrm{y}+3 \mathrm{z}=0$
$\Rightarrow \frac{y}{z}=-3$
$x+\frac{y}{z}=-3$
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